Question

A particle executing $SHM$ with time period $T$ and amplitude $A$. The mean velocity of the particle averaged over quarter oscillation is

• A. $\frac{A}{4T}$
• B. $\frac{2A}{T}$
• C. $\frac{3A}{T}$
• D. $\frac{4A}{T}$

Answer 1

Answer: (D)

Let the displacement of the particle executing $SHM$ at any instant $t$ is , $x=Asin\omega t$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}Asin\omega t=Awcos\omega t$
The mean velocity of the particle averaged over quarter oscillation is
$(v{)}_{0\to T/4}=\frac{\underset{0}{\overset{T/4}{\int }}vdt}{\underset{0}{\overset{T/4}{\int }}dt}=\frac{\underset{0}{\overset{T/4}{\int }}A\omega cos\omega tdt}{T/4}$
$\frac{\frac{A\omega }{\omega }{\left[sin\omega t\right]}_0^{T/4}}{T/4}$
$=\frac{A{\left[sin\omega t\right]}_0^{T/4}}{T/4}=\frac{4A}{T}$